I don't know what that line above the question is. Thanks in Advance.

Download 1.3-Complex_Numbers_bad.mw

I'm trying to get my problems in standard form  a + bi . Questions 19 - 22 are wrong.

How can I display the symbol (blue solid circle) and the line (blue line) together in the legend box (( from   to ))?

Download Plot1.mw

As a Maple beginner, I am now interested in symbolic calculations in Maple. As before, I set a problem from a subject area that interests me in order to learn from professional answers.

Determine all regular square (n;n) matrices (determinant not equal to zero) that are commutable with every regular (n;n) matrix with respect to matrix multiplication.

(I know the solution from long ago.)

Example code

printlevel :=1     

for indx1 from 1 by 1 to  3  do  

f[indx1] :=indx1;  

 end do;

This prints
f₁:=1
f₂:=2
f₃:=3

How do I use the "save" command to save exactly the loop's results above to a file so that I can read the file later and execute it in another  maple worksheet.

The maple manual https://www.maplesoft.com/support/help/Maple/view.aspx? 

for "save"   contains no examples, and definately not how to save results of a loop using the save command. e.g. how do you append  with a file using the "save" command ?

Note: I dont need any help with reading the results, from the saved file, my question is only about writing the results with "save" command. The "save" command gives me the best results for reading files back into a speadsheet, and text file save routines just gives me ascii garbage and not the exact results in executable maple format as "save" does, saving exacltly what you see on screen. Therefore text save routines are useless to me.

I have a file TEST.m. How can I make it so that every time I start Maple, all the subprograms in the TEST.m file will run first? Then I just need to type the function with(TEST): sumpro(2,3,4) to get the result 9. I copied the TEST.m file into Maple's lib directory, but it doesn't run after starting Maple.

I just need to type sumvip(2, 3, 4) to get the result, but Maple doesn't understand it.

Please help.

Download TEST.mw

2024-12-20_Q_simplification_Question.mw
Solve the general cubic. Apply values and simplify. 

Could someone show how Maple simplifies to the value of X=3? I tried doing it manually and I could not figure it out. 

Also is there a Help assistant to see the setps?

Download 2024-12-20_Q_simplification_Question.mw

In the decimal system, we are looking for all natural numbers with at most six digits that only swap the order of the digits when multiplied by 2, 3, ..., 6.

In a plane, equilateral triangles D(i) with side lengths a(i)= 2*i−1, i = 1; 2; 3; ... are arranged along a straight line g in such a way that the "right" corner point of triangle D(k) coincides with the "left" corner point of triangle D(k+1) and that the third corner points all lie in the same half-plane generated by g. Determine the curve/function on which the third corner points lie!

I am very happy to announce the first public release of a project which I have been working on for the last couple of years.

NODEMaple consists of a set of Maple workbooks and a library for structural design based on the Eurocode.

Currently the main development of the workbooks is focused on "Eurocode 5: Design of timber structures" with the Norwegian Annex.

This software has been made public in the hope of that it might be useful for other structural designers, professionals as well as students. Everyone interested is very Welcome to contribute to this project. The code is published under the GPLv3 license.

For more information see https://github.com/Anthrazit68/NODEMaple.

Good afternoon.

I have a differential equation of non-integer degree and would like to know if it is possible to express a solution in terms of elementary or special-functions for certain values of the exponent, n>0.

For this equation, Maple provides an analytical solution for the exponent values n=0 and n=1, otherwise, there is no solution returned. I am particularly interested in the cases where n=1/2, 3/2, 2, 5/2, and 3

I am hoping that someone can help me resolve this - if a closed-form solution is not possible, then a numerical solution would also be welcome.

I have provided the details in the attached worksheet.

Thanks for reading!

MaplePrimes_Dec_19.mw

Given the center x1 of a circle in R^2 with radius d12, and a point p2 on the circle, so that d12=||p2-x1||, denote the points on the line segment from x1 to p2 as x1(t) = x1+t*v12, with t=0..d12, and v12 =( p2-x1)/d12.  I want to animate the points x1(t) moving along the line segment from x1 to p2 and the corresponding circles of decreasing radius, with center x(t) and radius d12-t, so that p2 remains on the circle.

I can animate the points along the line segment from x1 to p2 using ‘style=point, symbol=solidcircle’.

I would like to use plottools-circle, to plot the circles. I have also tried the following type commands for the circles of decreasing radius.

Plot([x1(1)+t*v12(1)+(d12-t)*cos(theta)*v12(1)+ (d12-t)*sin(theta)*u12(1), x1(2)+ +t*v12(2)+(d12-t)*d12*cos(theta)*v12(2)+(d12-t)*sin(theta)*u12(2), theta=0..2*PI]

where u12 is a unit vector orthogonal to v12.

I have not been able to combine the two plots into an animation. Thank you

On considère une ellipse x^2/a^2+y^2/b^2-1=0 et 2 sommets de cette ellipse A(a,0) et B(0,b). On imagine une hyperbole équilatère variable passant par les points O, A et B. Cette courbe rencontre l'ellipse en 2 autres points A1 et B1. Montrer que la droite A1B1 passe par un point fixe. Même avec l'intelligence artificielle, je ne parviens pas à résoudre ce problème. Pourriez-vous d'aider. Merci.

Machine translation by moderator:

We consider an ellipse x^2/a^2+y^2/b^2-1=0 and 2 vertices of this ellipse A(a,0) and B(0,b). We imagine a variable equilateral hyperbola passing through the points O, A and B. This curve meets the ellipse at 2 other points A1 and B1. Show that the line A1B1 passes through a fixed point. Even with artificial intelligence, I can't solve this problem. Could you help. Thank you.

I didn't put it in the title, but of course this is a post about Advent of Code, in particular Days 16 and 18 which feature a perenial favorite type of problem: finding shortest paths in mazes.

Your input for these is always a maze given as an ascii map.  Like so:

###############
#.......#....E#
#.#.###.#.###.#
#.....#.#...#.#
#.###.#####.#.#
#.#....#....#.#
#.#.#####.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S......#.#...#
###############

There's lots of ways to import one of these into Maple and then solve the maze, but I am to highlight how to do it with GraphTheory.  I am going to start with a GridGraph and then remove the walls in order to leave a just the vertices that represent the paths:

with(StringTools): with(GraphTheory):
maze:=
"###############
#.......#....E#
#.#.###.#.###.#
#.....#.#...#.#
#.###.#####.#.#
#.#....#....#.#
#.#.#####.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S......#.#...#
###############
":
mazelines := (Split(Trim(maze), "\n")):
sgrid := ListTools:-Reverse((map(Explode, mazelines)) ):
m,n := nops(sgrid), nops(sgrid[1]);
tgrid := table([seq(seq([i,j]=sgrid[i,j],i=1..m),j=1..n)]):
start := lhs(select(e->rhs(e)="S", [entries(tgrid,'pairs')])[]);
finish := lhs(select(e->rhs(e)="E", [entries(tgrid,'pairs')])[]);

Now the maze is stored in the table tgrid, and it is easy to find the walls and paths.  In a GridGraph the vertices are labeled with their coordinates as "x,y" and so we rewrite our list of paths in that form, so we can create the induced subgraph of the Grid that includes only those vertices.

(walls,paths) := selectremove(e->rhs(e)="#", [entries(tgrid, 'pairs')]):
paths := map(s->sprintf("%d,%d",lhs(s)[]), paths):
H := SpecialGraphs:-GridGraph(m,n);
G := InducedSubgraph(H, paths);

We can use StyleVertex to highlight the start and finish.

StyleVertex(G, sprintf("%d,%d",start[]), color="LimeGreen");
StyleVertex(G, sprintf("%d,%d",finish[]), color="Red");

plots:-display(<
DrawGraph(H, stylesheet=[vertexshape="square", vertexborder=false, vertexcolor="Black"], showlabels=false) | 
DrawGraph(G, stylesheet=[vertexshape="square", vertexborder=false, vertexcolor="Black"], showlabels=false)>);

(I omitted a step where I set the vertex locations of the maze grid, you can see that in the attached worksheet)

Now finding a path through the maze is as easy as calling GraphTheory:-ShortestPath

sp := ShortestPath(G, sprintf("%d,%d",start[]), sprintf("%d,%d",finish[]) ):

StyleVertex(G, sp[2..-2], color="Orange");
StyleEdge(G, [seq({sp[i],sp[i+1]}, i=1..nops(sp)-1)], color="Orange");
DrawGraph(G, stylesheet=[vertexshape="square", vertexpadding=10, vertexborder=false,
             vertexcolor="Black"],  showlabels=false, size=[800,800]);

Now, Advent of Code seldom gives you a completely simple maze like this, often these is a twist like having to calculate the costs of turns seperately from the cost of steps, or each direction or position has a seperate cost associated with it.

For example, Day 16 has us starting facing east, and then turns cost 1000, while moving forward costs 1. That sort of problem is no longer exactly a maze, instead of the vertices being representing an "x,y" position, instead you increase the number of vertices by a factor of 4, so that you have a vertex for every position and orientation "x,y,o" with edges of weight 1 between adjacent vertices with the same orientation and edges of wieght 1000 to connect "x,y,N" to "x,y,E" and "x,y,W" e.g.  In that sort of weighted graph, we can use GraphTheory:-DijkstrasAlgorithm to find the shortest path and it's weighted cost.

In this code, we expand our list of maze locations with directions, and the use the grid table to generate a list of weighted edges:

dtable := table([0=[0,1], 1=[1,0], 2=[0,-1], 3=[-1,0]]):
dname := table([0="N",1="E",2="S",3="W"]):
dpaths := map(s->local d;seq(cat(s,",",d), d in ["N","E","S","W"]), paths):

edges := NULL:
for i from 1 to m do for j from 1 to n do
    if tgrid[[i,j]] = "#" then next; end if;
    for d from 0 to 3 do
        dir := dtable[d];
        if tgrid[[i,j]+dir] <> "#" then
            edges := edges, [{cat("",i,",",j,",",dname[d]), cat("",i+dir[1],",",j+dir[2],",",dname[d])},1];
        end if;
        edges := edges, [{cat("",i,",",j,",",dname[d]), cat("",i,",",j,",",dname[d+1 mod 4])}, 1000],
                 [{cat("",i,",",j,",",dname[d]), cat("",i,",",j,",",dname[d-1 mod 4])}, 1000];
    end do;
end do; end do:

Gd := Graph(dpaths,weighted,{edges});

Once that is done, it's a simple matter of calling Dijkstra's Algorithm on the graph, but notice that we can reach the finsh while traveling north or east, so we need to find the sortest path to both (you can pass a list of vertices to Dijkstra, and it will efficiently calculate paths to all of them), and select the smaller of the two:

spds := DijkstrasAlgorithm(Gd, cat("",start[1],",",start[2],",E"), 
    [cat("",finish[1],",",finish[2],",N"), cat("",finish[1],",",finish[2],",E")] , 
    distance):
i := min[index](map2(op,2,spds)):
spd := spds[i];

spd := [["2,2,E", "3,2,E", "4,2,E", "4,2,N", "4,3,N", "4,4,N", "4,5,N", "4,6,N", "4,6,E", "5,6,E",
 "6,6,E", "7,6,E", "8,6,E", "8,6,N", "8,7,N", "8,8,N", "8,9,N", "8,10,N", "8,11,N", "8,12,N", 
"8,12,W", "7,12,W", "6,12,W", "5,12,W", "4,12,W", "3,12,W", "2,12,W", "2,12,N", "2,13,N", 
"2,14,N", "2,14,E", "3,14,E", "4,14,E", "5,14,E", "6,14,E", "7,14,E", "8,14,E", "9,14,E", 
"10,14,E", "11,14,E", "12,14,E", "13,14,E", "14,14,E"], 6036]

We can then plot to compare this to the unweighted shortest path:

dsp := ListTools:-MakeUnique( map(s->s[1..-3], spd[1]) );
StyleVertex(G, dsp[2..-2], color="DarkBlue");
StyleEdge(G, [seq({dsp[i],dsp[i+1]}, i=1..nops(dsp)-1)], color="DarkBlue");

DrawGraph(G, stylesheet=[vertexshape="square", vertexpadding=10,
             vertexborder=false, vertexcolor="Black"],  showlabels=false,
          size=[800,800]);

And you can see it's a path that requires more steps, but definitely uses fewer turns if we start facing east/right (6 vs. 9):

I hope this has given you a little bit of a flavor of how to use GraphTheory commands to solve path finding problems.  Like with the second part here, usually the biggest challenge is figuring out how to encode and construct a graph that represents your problem.  Then the actual commands to solve it, are easy. You can see all the code, and a couple steps I left out from above in this worksheet: Mazeblog.mw

And just for fun, here's a Maple workbook that imports a maze from an image and solves it: MazeFromImage.maple